Question

A playground merry-go-round has a radius R and a rotational inertia I. When the merry-go-round is at rest, a child with mass m runs with speed v along a line tangent to the rim and jumps on. The angular velocity of the merry-go-round is then

Answer #1

Initial angular momentum of the child + merry-go-round system
with respect to the center of the merry-go-round is,

Li = m * v * R

Once the child is at the rim of the merry-go-round, the moment
of inertia changes,

Final moment of inertia, If = I + m * R2

Consider that the merry-go-round rotates with an angular velocity

Final angular momentum, Lf = If *

= (I + m * R2) *

Using the conservation of angular momentum, Li = Lf

m * v * R = (I + m * R2) *

= **(m * v * R) / (I + m * R2)**

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3)...

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inertia I = 182 kg-m2 is spinning with an initial
angular speed of ω = 1.4 rad/s in the counter clockwise direection
when viewed from above. A person with mass m = 71 kg and velocity v
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(8%)
Problem 17: A merry-go-round is a
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